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- KingGeorge

[SOLVED] Here's a cute predicate logic problem.
Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).

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- KingGeorge

- jamiebookeater

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- experimentX

is xyzw a combination of digit or multiplication of digit??

- KingGeorge

multiplication

- experimentX

lol .. didn't realize they were complex

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- KingGeorge

It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.

- KingGeorge

Just prove/disprove it for \(\mathbb{R}\) and it will become clear why it doesn't matter if they're complex numbers.

- KingGeorge

Hint below in white LaTeX\[\color{white}{\text{There is a certain value for y that you should choose to prove it.}}\]

- experimentX

\[ \color{white}{\text{LOL i didn't know that!!.}} \]

- KingGeorge

Another, probably more helpful hint, below in white LaTeX\[\color{white}{\text{What value of y can you choose so that xyzw is constant?}}\]

- anonymous

Choose y=0, then choose w=-(x+z).

- KingGeorge

Bingo.

- experimentX

But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??

- KingGeorge

\(\exists\, y\) means that we can choose a specific value for y. It's a variable, but one that we have complete control over.

- experimentX

then it must be false ... i guess so!!

- KingGeorge

The "there exists" only means that at least one possible value of y where this is true. One value that we know works is \(y=0\). This means that there does exists at least one value for \(y\) that makes the statement true, so we have proven that the original statement is true.

- experimentX

does't y=0 make statement untrue??
xyzw=0, for y=0

- anonymous

Right, both the sum and the multiplication end up equaling zero because you set w=-(x+z).

- KingGeorge

If \(y\) is 0, and \(x=-(x+z)\) we have \[x+0+z+(-(x+z))=x+z-x-z=0\]and \[x\cdot0\cdot z\cdot w=0\]So that means that \[x+y+z+w=xyzw\]

- experimentX

Oh ... stupid me ... i never considered the negative values!!

- KingGeorge

Hope everyone enjoyed this little (unintuitive) problem. Good night.

- experimentX

thanks!! i'll try to think broader way next time!!

- KingGeorge

I'll try to get a problem just for you next time then. Any preference about the subject?

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